15- Solved problem 6-19-4 for local buckling of column-2010.

Last Updated on March 4, 2026 by Maged kamel

Solved problem 6-19-4 for local buckling of column-2010.

Solved problem 6-19-4-estimate area and inertia values of the given section.

The next slide image shows a brief introduction to the content of this post.

 What is the main chapter for the design of compression members? The main chapter is Chapter E, and the relevant sections for the built-up section are E6 and E7 for slender members.

 Determine the nominal axial strength for the Non-standard shape of Figure 6.19.4 for an effective length kl=8 feet, with fy used of 100 ksi. The I  section is composed of three plates; the upper plate breadth is 10 inches. The thickness is 1/2 inch.

The same dimension applies to the lower plate as well. The middle plate has a height of 11 inches and a width of 1/4 inch; thus, the overall height is 12 inches.
Given k*L is 8 feet. Since this section is a built-up section, we cannot obtain information from any tables. The following slide image shows the calculation of Ix, the moment of inertia about the x-x axis; the Ix value is 358.56 in. 4.

For local buckling, we have to estimate Ix and Iy. The value of rx is much higher than ry, the minor direction will govern buckling.

We will estimate the Iy, since buckling will be about the minor axis. the area of section equals 12.75 inch2, the moment of inertia about Y axis equals 83.35 inch4. The radius of gyration equals 2.56 inches. Please refer to the following slide image for more details.

Determine which direction controls the design.

We can find the value of the radius of gyration about the x-x, which is sqrt(358.56/12.75)=5.30 inches. Find the values of (Kl/rx) and Kl/ry, For Kl/rx=96/5.3=11.62.

While Kl/ry=(96/2.56)=37.50. Since Kl/ry is bigger than Kl/rx, the buckling about the Y-axis will control the design.

Check b/t for unstiffened part-Flange and reduction factor Qs for unstiffened part.

The ratio of b/t is the ratio of half the flange length to the flange, upper plate thickness, and equals (10/2/0.50) = 10. Check this ratio with B/t as given by E7-7 Equation, which gives this relation, b/t <=0.64*sqrt(E*kc/Fy).

The kc value equals 4/sqrt(h/tw)=4/sqrt(11/4)=0.603, since hw=11″ and web thickness, middle plate=0.25 inches. The required b/t =0.64*sqrt(29000*0.603/100)=8.46.

Since b/t of the section is bigger than 8.46, Qs is not equal to one. We need to proceed to the E7-8 equation to find the equation for Qs.

Estimate b/t based on the value of 1.17*sqrt(E*kc/Fy), which equals(1.17*sqrt(29000*0.603/100)=15.47.

The actual b/t,10, is bigger than 8.46 but less than 15.47. The Qs equation 1.145-0.65*(b/t)*sqrt(Fy/E*kc)=1.145-10*sqrt(100/29000*0.603)=0.923. Please refer to the following slide image.

For the stiffened part, the ratio of (h/tw)=11/0.25=44.

Find the affection b value and the effective area based on Q=1.0.

Now, we will proceed to calculate Qa. For the slender stiffened element, h/tw=11/0.25=44 will be compared with (1.49*sqrt(E/Fy),  which gives the value of  1.49*sqrt(29000/100)=25.40.

Our h/tw for the section > 25.4: since h/tw > λr, Qa is <1, since h/tw is > λr. Then the modified h will be based on E7-17. But we need to find the value of the stress f in the equation; consider Qa = 1.0.

Find the Euler stress and critical stress for the column.

The column is inelastic since Kl/r y value which equals 37.50, is less than 4.71*sqrt(E/Q*Fy) which equals 80.208, The facr=0.658^(Q*fy/Fe)*(Q*Fy). The Euler stress=Pi^2*E/(Kl/r)^=204 ksi,

First, consider Qa=1, our calculated Qs=0.923, then q=1*0.923=0.923,for the first trial, Qfy=0.923*100=92.3 ksi, Fcr=Qfy*(0.658^Qfy/fE)=92.3* 0.658^(92.3/204)=76.38 ksi.

The new graph for the short column is shown as a dotted line.
Take the fcr value, which is 76.38 ksi, back into equation E 7-17 as follows.

Find the effective b-value and the effective area based on Q=0.923.

Then be=1.92*(tw)*sqrt (E/f)*(1-(0.34/bt*Sqrt(E/f), tw=0.25 *sqrt(29000/f), the trial f=Qa*Fcr as estimated earlier, the trial f=0.923*76.28, Trial f=0.923*76.38=70.50 ksi, the term 1-(0.34/bt*Sqrt(E/f)=1-(0.34/11/0.25*sqrt(29000/70.5)=8.209 inches<11 inches which is the web height.

The effective area=Ag-hw*hw+be*tw=12.75-11*0.25+8.209*0.25=12.05 inch2. Next, we will estimate the Qa value.

Reestimate the Qa=Aeff /Agr=12.05/12.75=0.945. For the new value of Qa. Qa=12.05/12.75=0.9452. So the Q new = Qs*Qa; the value of Qa = 0.945 while Qs = 0.923.

Finally, Q new = 0.923*0.945 = 0.8722, back to Fcr = 72.94 ksi. Again, multiply by Q = 0.8722 to get another value of stress f=0.8722*72.94=63.60 ksi.

Multiply by Q to get the new f, which is 0.8722*72.9=63.6 ksi.

Find the affection b value and the effective area based on Q=0.8722.

Evaluate the modified value of (be ) for the stiffened web, be=1.92*(0.25)*sqrt(29000/63.6)* by the other bracket—finally, hw eff=8.557 inches.

Recalculate the A eff=10*0.5*2+8.557*0.25, Aeff =12.14 inch2. The corresponding effective area is 12.14 in^2.

Find the effective b-value and the effective area based on Q=0.872.

Qa new =A eff/Ag=12.14/12.75=0.945, To get the  Q=Qs*Qa=0.923*0.945= 0.872,  fcr recalculated with the new Q value, Fcr=72.94 ksi. Multiply by Q to get the new f, which is 0.872*72.94=63.6 ksi. Evaluate the modified value of (be ) for the stiffened web, be=1.92*(0.25)*sqrt(29000/63.6)* by the other bracket—finally, hw eff=8.557 inches, same as estimated earlier.

Solved problem 6-19-4-estimate the nominal load.

The A eff=12.75-(11*0.25)+(8.557*0.25)=12.14 inch2. The Final Qa value=12.14/12.75=0.952. The final Q value=0.923*0.952=0.879. Estimate fcr=0.658^(0.879*100/204)*87.90=73.0 ksi. there Nominal load =fcr *Ag=73*12.75=931 kips.

In this LRFD-only example, φc = 0.9, and φc*Pn = 0.9*931 = 838 kips. This is the end of our solved problem 6-19-4. Thanks a lot.

The PDF file for this post can be viewed and downloaded from the following link.

For a good A Beginner’s Guide to the Steel Construction Manual, 14^th ed. Chapter 7 – Concentrically Loaded Compression Members.

For a good A Beginner’s Guide to the Steel Construction Manual, 15^th ed. Chapter 7 – Concentrically Loaded Compression Members.

For a good A Beginner’s Guide to the Steel Construction Manual, 16^th ed. Chapter 7 – Concentrically Loaded Compression Members.

For the next post, A Solved Problem 5-10 For the Available Strength.